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Question

Physics Question on Alternating current

A small town with a demand of 800kW800\,kW of electric power at 220V220\,V is situated 15km15\,km away from an electric plant generating power at 440V440\,V. The resistance of the two wire line carrying power is 0.5Ω0.5\,\Omega. per kmkm. The town, gets power from the line through a 4000220V4000-220\,V step down transformer at a substation in the town. The line power loss in the form of heat is

A

400kW400\,kW

B

600kW600\,kW

C

300kW300\,kW

D

800kW800\,kW

Answer

600kW600\,kW

Explanation

Solution

Here, P=800kW=800×103WP = 800\,kW = 800 \times 10^3\, W Total resistance of two wire line R=2×15×0.5=15ΩR = 2 \times 15 \times 0.5 = 15\,\Omega As supply is through 4000220V4000-220\,V transformer Vrms=4000V\therefore V_{rms}=4000\,V Irms=PVrms\therefore I_{rms}=\frac{P}{V_{rms}} =800×1034000=\frac{800\times10^{3}}{4000} =200A=200\,A Line power loss =Irms2R=(200)2×15=I^{2}_{rms}\,R=\left(200\right)^{2}\times 15 =60×104W=600kW=60\times 10^{4}\,W=600\,kW